Solve 10h^2-9h-9

$$10 h ^ { 2 } - 9 h - 9$$

Factor

$$\left(2h-3\right)\left(5h+3\right)$$

Solution Steps

Factor the expression by grouping. First, the expression needs to be rewritten as 10h^{2}+ah+bh-9. To find a and b, set up a system to be solved.

$$a+b=-9 ab=10\left(-9\right)=-90$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -90.

$$1,-90 2,-45 3,-30 5,-18 6,-15 9,-10$$

Calculate the sum for each pair.

$$1-90=-89 2-45=-43 3-30=-27 5-18=-13 6-15=-9 9-10=-1$$

The solution is the pair that gives sum -9.

$$a=-15 b=6$$

Rewrite 10h^{2}-9h-9 as \left(10h^{2}-15h\right)+\left(6h-9\right).

$$\left(10h^{2}-15h\right)+\left(6h-9\right)$$

Factor out 5h in the first and 3 in the second group.

$$5h\left(2h-3\right)+3\left(2h-3\right)$$

Factor out common term 2h-3 by using distributive property.

$$\left(2h-3\right)\left(5h+3\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$10h^{2}-9h-9=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$h=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 10\left(-9\right)}}{2\times 10}$$

Square -9.

$$h=\frac{-\left(-9\right)±\sqrt{81-4\times 10\left(-9\right)}}{2\times 10}$$

Multiply -4 times 10.

$$h=\frac{-\left(-9\right)±\sqrt{81-40\left(-9\right)}}{2\times 10}$$

Multiply -40 times -9.

$$h=\frac{-\left(-9\right)±\sqrt{81+360}}{2\times 10}$$

Add 81 to 360.

$$h=\frac{-\left(-9\right)±\sqrt{441}}{2\times 10}$$

Take the square root of 441.

$$h=\frac{-\left(-9\right)±21}{2\times 10}$$

The opposite of -9 is 9.

$$h=\frac{9±21}{2\times 10}$$

Multiply 2 times 10.

$$h=\frac{9±21}{20}$$

Now solve the equation h=\frac{9±21}{20} when ± is plus. Add 9 to 21.

$$h=\frac{30}{20}$$

Reduce the fraction \frac{30}{20} to lowest terms by extracting and canceling out 10.

$$h=\frac{3}{2}$$

Now solve the equation h=\frac{9±21}{20} when ± is minus. Subtract 21 from 9.

$$h=-\frac{12}{20}$$

Reduce the fraction \frac{-12}{20} to lowest terms by extracting and canceling out 4.

$$h=-\frac{3}{5}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{3}{2} for x_{1} and -\frac{3}{5} for x_{2}.

$$10h^{2}-9h-9=10\left(h-\frac{3}{2}\right)\left(h-\left(-\frac{3}{5}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$10h^{2}-9h-9=10\left(h-\frac{3}{2}\right)\left(h+\frac{3}{5}\right)$$

Subtract \frac{3}{2} from h by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$10h^{2}-9h-9=10\times \frac{2h-3}{2}\left(h+\frac{3}{5}\right)$$

Add \frac{3}{5} to h by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$10h^{2}-9h-9=10\times \frac{2h-3}{2}\times \frac{5h+3}{5}$$

Multiply \frac{2h-3}{2} times \frac{5h+3}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$10h^{2}-9h-9=10\times \frac{\left(2h-3\right)\left(5h+3\right)}{2\times 5}$$

Multiply 2 times 5.

$$10h^{2}-9h-9=10\times \frac{\left(2h-3\right)\left(5h+3\right)}{10}$$

Cancel out 10, the greatest common factor in 10 and 10.

$$10h^{2}-9h-9=\left(2h-3\right)\left(5h+3\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.This is achieved by dividing both sides of the equation by 10

$$x ^ 2 -\frac{9}{10}x -\frac{9}{10} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = \frac{9}{10} rs = -\frac{9}{10}$$

Two numbers r and s sum up to \frac{9}{10} exactly when the average of the two numbers is \frac{1}{2}*\frac{9}{10} = \frac{9}{20}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{9}{20} - u s = \frac{9}{20} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{9}{10}

$$(\frac{9}{20} - u) (\frac{9}{20} + u) = -\frac{9}{10}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{81}{400} - u^2 = -\frac{9}{10}$$

Simplify the expression by subtracting \frac{81}{400} on both sides

$$-u^2 = -\frac{9}{10}-\frac{81}{400} = -\frac{441}{400}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{441}{400} u = \pm\sqrt{\frac{441}{400}} = \pm \frac{21}{20}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =\frac{9}{20} - \frac{21}{20} = -0.600 s = \frac{9}{20} + \frac{21}{20} = 1.500$$

Evaluate

$$\left(2h-3\right)\left(5h+3\right)$$